Synchronization of delayed coupled neurons in presence of inhomogeneity

Abstract

In principle, two directly coupled limit cycle oscillators can overcome mismatch in intrinsic rates and match their frequencies, but zero phase lag synchronization is just achievable in the limit of zero mismatch, i.e., with identical oscillators. Delay in communication, on the other hand, can exert phase shift in the activity of the coupled oscillators. In this study, we address the question of how phase locked, and in particular zero phase lag synchronization, can be achieved for a heterogeneous system of two delayed coupled neurons. We have analytically studied the possibility of inphase synchronization and near inphase synchronization when the neurons are not identical or the connections are not exactly symmetric. We have shown that while any single source of inhomogeneity can violate isochronous synchrony, multiple sources of inhomogeneity can compensate for each other and maintain synchrony. Numeric studies on biologically plausible models also support the analytic results.

Appendix: The model neurons

Wang–Buzsaki (WB) model

The steady-state activation m _∞ and the rate equation for the inactivation variable h in the expression for sodium current and rate equation for the activation variable n in the expression for potassium current are given respectively as follow:

$$\begin{array}{@{}rcl@{}} m&=&m_{\infty} (V) = \dfrac{\alpha_{m} (V)}{\alpha_{m} (V) + \beta_{m} (V)},\\ \dfrac{dh}{dt}&=&\phi [\alpha _{h} (V) (1 - h) - \beta _{h} (V) h],\\ \dfrac{dn}{dt}&=&\phi [\alpha _{n} (V) (1 - n) - \beta _{n} (V) n]. \end{array} $$

(20)

The rate constants for m _∞, h, and n are:

$$ \begin{array}{l} \alpha_{m} (V) = -0.1 (V + 35)/(\exp (-0.1(V + 35)) - 1),\\ \beta_{m} (V) = 4 \exp (- (V +60)/18), \end{array} $$

$$ \begin{array}{l} \alpha_{h} (V) = 0.07 \exp (- (V + 58))/20),\\ \beta_{h} (V) = 1/ (\exp (-0.1 (V +28) + 1), \end{array} $$

(21)

$$ \begin{array}{l} \alpha_{n} (V) = -0.01 (V + 34)/(\exp (-0.1(V + 34)) - 1),\\ \beta_{n} (V) = 0.125 \exp (- (V +44)/80). \end{array} $$

Hodgkin–Huxley (HH) model

The rate equations for the activation variable m and inactivation variable h of the sodium expression, and n, activation variable of potassium, obey the differential equations:

$$ \begin{array}{l} \dfrac{dm}{dt} = [\alpha _{m} (V) (1 - m) - \beta _{m} (V) m],\\\\ \dfrac{dh}{dt} = [\alpha _{h} (V) (1 - h) - \beta _{h} (V) h],\\\\ \dfrac{dn}{dt} = [\alpha _{n} (V) (1 - n) - \beta _{n} (V) n].\\ \end{array} $$

(22)

The rate constants for m, h, and n are:

$$ \begin{array}{l} \alpha _{m} (V) = {{(2.5 - 0.1V)}}/{{(\exp{(2.5 - 0.1V)} - 1)}},\\ \beta _{m} (V) = 4\exp{ {(-V}/{{18)}}},\end{array} $$

$$ \begin{array}{l} \alpha _{h} (V) = 0.07\exp{ {(-V}/{{20)}}},\\ \beta _{h} (V) = {1}/{{(\exp{(3 - 0.1V)} + 1)}}, \end{array} $$

(23)

$$ \begin{array}{l} \alpha _{n} (V) ={{(0.1 - 0.01V)}}/{{(\exp{(1 - 0.1V)} - 1)}},\\ \beta _{n} (V) = 0.125\exp{ {(-V}/{{80)}}}. \end{array} $$